\(\frac{1}{M}=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{59.60}{2}}\)

\(\frac{1}{M}=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}\)

\(\frac{1}{M}=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{59}-\frac{1}{60}\right)\)

\(\frac{1}{M}=\frac{2}{3}-\frac{2}{60}< \frac{2}{3}\)

-theo t đề là M chứ ko phải 1/M 

\(M=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+59}\\ =\dfrac{1}{\dfrac{3\cdot4}{2}}+\dfrac{1}{\dfrac{4\cdot5}{2}}+...+\dfrac{1}{\dfrac{59\cdot60}{2}}\\ =\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{59\cdot60}\\ =2\left(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{59\cdot60}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\right)\\ =2\cdot\dfrac{19}{60}\\ =\dfrac{38}{60}< \dfrac{40}{60}=\dfrac{2}{3}\)

1.A = 2¹ + 2² + 2³ + 2⁴ + ... + 2⁵⁹ + 2⁶⁰

Xét .dãy số: 1; 2; 3; 4; .... 59; 60 Dãy số này có 60 số hạng vậy A có 60 hạng tử.

vì 60 : 2 = 30 nên nhóm hai số hạng liên tiếp của A vào một nhóm thì ta được:

A = (2¹ + 2²) + (2³ + 2⁴) +...+ (2⁵⁹ + 2⁶⁰)

A = 2.(1 + 2) + 2³.(1 +2) +...+ 2⁵⁹.(1 +2)

A =2.3 + 2³.3  + ... + 2⁵⁹.3

A =3.( 2 + 2³+...+ 2⁵⁹)

Vì 3 ⋮ 3 nên A = 3.(2 + 2³ + ... + 2⁵⁹)⋮3 (đpcm)

áp dụng công thức là ra :))))

2, M = 3ⁿ⁺³ + 3ⁿ⁺¹ + 2ⁿ⁺³ + 2ⁿ⁺² ⋮ 6

   M = 3ⁿ⁺¹.(3² + 1) + 2ⁿ⁺².(2 + 1) 

    M = 3ⁿ.3.(9 + 1) + 2ⁿ⁺¹.2 . 3

    M = 3ⁿ.30 + 2ⁿ⁺¹.6

   M = 6.(3ⁿ.5 + 2ⁿ⁺¹)

   Vì 6 ⋮ 6 nên M = 6.(3ⁿ.5+ 2ⁿ⁺¹) ⋮ 6 (đpcm)

Ta có : \(\frac{1}{m}=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}\)

\(\Rightarrow m=\frac{30}{19}>\frac{2}{3}\)

\(Tac\text{ó}:\frac{1}{m}=\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{59.60}=2\left(\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{59}-\frac{1}{60}\right)\)

\(=>2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}\\ =>m=\frac{30}{19}>\frac{2}{3}\)

\(\frac{1}{M}=\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+...+\frac{1}{1+2+3+...+59}\)

\(\frac{1}{M}=\frac{1}{3\left(1+3\right):2}+\frac{1}{4\left(1+4\right):2}+\frac{1}{5\left(1+5\right):2}+...+\frac{1}{59\left(1+59\right):2}\)

\(\frac{1}{M}=\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{59.60}\)

\(\frac{1}{M}=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)

\(\frac{1}{M}=2\left(\frac{1}{3}-\frac{1}{60}\right)\)

\(\frac{1}{M}=\frac{1}{2}.\frac{19}{60}\)

\(\frac{1}{M}=\frac{19}{120}\)

\(M=\frac{120}{19}>\frac{2}{3}\left(đpcm\right)\)

chuẩn men

\(M=\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+59}\)

\(M=\frac{1}{\frac{3.\left(3+1\right)}{2}}+\frac{1}{\frac{4.\left(4+1\right)}{2}}+\frac{1}{\frac{5.\left(5+1\right)}{2}}+...+\frac{1}{\frac{59.\left(59+1\right)}{2}}\)

\(M=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+\frac{1}{\frac{5.6}{2}}+...+\frac{1}{\frac{59.60}{2}}\)

\(M=\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{59.60}\)

\(M=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{59.60}\right)\)

\(M=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)

\(M=2.\left(\frac{1}{3}-\frac{1}{60}\right)\)

\(M< 2.\frac{1}{3}\)

\(M< \frac{2}{3}\)